Namespaces
Variants
Actions

Difference between revisions of "Urysohn lemma"

From Encyclopedia of Mathematics
Jump to: navigation, search
(Importing text file)
 
(TeX)
 
Line 1: Line 1:
For any two disjoint closed sets <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958801.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958802.png" /> of a [[Normal space|normal space]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958803.png" /> there exists a real-valued function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958804.png" />, continuous at all points, taking the value <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958805.png" /> at all points of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958806.png" />, the value 1 at all points of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958807.png" /> and for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958808.png" /> satisfying the inequality <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u0958809.png" />. This lemma expresses a condition which is not only necessary but also sufficient for a <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u09588010.png" />-space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095880/u09588011.png" /> to be normal (cf. also [[Separation axiom|Separation axiom]]; [[Urysohn–Brouwer lemma|Urysohn–Brouwer lemma]]).
+
{{TEX|done}}
 +
For any two disjoint closed sets $A$ and $B$ of a [[Normal space|normal space]] $X$ there exists a real-valued function $f$, continuous at all points, taking the value $0$ at all points of $A$, the value 1 at all points of $B$ and for all $x\in X$ satisfying the inequality $0\leq f(x)\leq1$. This lemma expresses a condition which is not only necessary but also sufficient for a $T_1$-space $X$ to be normal (cf. also [[Separation axiom|Separation axiom]]; [[Urysohn–Brouwer lemma|Urysohn–Brouwer lemma]]).
  
  

Latest revision as of 16:02, 19 April 2014

For any two disjoint closed sets $A$ and $B$ of a normal space $X$ there exists a real-valued function $f$, continuous at all points, taking the value $0$ at all points of $A$, the value 1 at all points of $B$ and for all $x\in X$ satisfying the inequality $0\leq f(x)\leq1$. This lemma expresses a condition which is not only necessary but also sufficient for a $T_1$-space $X$ to be normal (cf. also Separation axiom; Urysohn–Brouwer lemma).


Comments

The phrase "Urysohn lemma" is sometimes also used to refer to the Urysohn metrization theorem.

References

[a1] A.V. Arkhangel'skii, V.I. Ponomarev, "Fundamentals of general topology: problems and exercises" , Reidel (1984) pp. 123–124 (Translated from Russian)
[a2] J.L. Kelley, "General topology" , v. Nostrand (1955) pp. 115
How to Cite This Entry:
Urysohn lemma. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Urysohn_lemma&oldid=15326
This article was adapted from an original article by P.S. Aleksandrov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article